(i) To find the equation of the tangent, we first differentiate the curve \(y = \sqrt{x^4 + 4x + 4}\).

The derivative is \(\frac{dy}{dx} = \left[ \frac{1}{2} (x^4 + 4x + 4)^{-\frac{1}{2}} \right] \times [4x^3 + 4]\).

At \(x = 0\), \(\frac{dy}{dx} = \frac{1}{2} \times 1 \times 4 = 2\).

The equation of the tangent is \(y - 2 = 2(x - 0)\), which simplifies to \(y - 2 = x\).

(ii) For the intersection, set \(x + 2 = \sqrt{x^4 + 4x + 4}\).

Squaring both sides gives \((x + 2)^2 = x^4 + 4x + 4\).

This simplifies to \(x^2 - x^4 = 0\), or \(x^2(1 - x^2) = 0\).

Thus, \(x = 0, \pm 1\).